MATHS 3 Marks Question

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man.
Accordingly, PA + PB = 10
We know that if a point moves in-plane in such a way that the sum of its distance from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.
Therefore, the path described by the man is an ellipse where the length of the major axis is 10m, while points A and B are the foci.
Taking the origin of the coordinate plane as the center of the ellipse, while taking the major axis along the x-axis,
The equation of the ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, where a is the semi-major axis.
Accordingly, 2a = 10 $\Rightarrow$ a = 5
Distance between the foci = 2ae = 2c = 8
$\Rightarrow$ c = 4
On using the relation, c = $\sqrt{a^{2}-b^{2}}$, we get,
4 = $\sqrt{25-b^{2}}$
$\Rightarrow$ 16 = 25 – b²
$\Rightarrow$ b² = 25 -1 6 = 9
$\Rightarrow$ b = 3
Put value of a and b in  $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
 $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ $\Rightarrow$ $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$.

Let l be the length of the rod and which at any position meet X-axis at A (a, 0) and also meets the Y-axis at B (0, b), therefore we have
l² = a² + b²
$\Rightarrow$ (12)² = a² + b² ...(i) [$\because$ l = 12]

Let P be the point on AB which is 3 cm from A and hence 9 cm from B.
This means that the point P divides AB in ratio 3 : 9 i.e., 1 : 3.
If P = (x, y), then by section formula, we have


(x, y) = $\left( \frac { 1 \times 0 + 3 \times a } { 1 + 3 } , \frac { 1 \times b + 3 \times 0 } { 1 + 3 } \right)$
$\Rightarrow$ (x, y) = $\left( \frac { 3 a } { 4 } , \frac { b } { 4 } \right)$
$\Rightarrow$ x = $\frac { 3 a } { 4 }$ , y = $\frac { b } { 4 }$ $\Rightarrow$ a = $\frac { 4 x } { 3 }$ and b = 4y
On putting the values of a and b in Equation (i), we get
144 = $\left( \frac { 4 x } { 3 } \right) ^ { 2 }$ + (4y)²
$\Rightarrow$ $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 9 }$ = 1
which is required equation.

$Since\ the\ circle\ passes\ through\ ( -1,1),\\ \therefore\ (-1-h)^2+(1-k)^2=r^2\\ \Rightarrow 1+h^2+2h+1+k^2-2k=r^2\\ \Rightarrow h^2+k^2+2h-2k+2=r^2......(2)$
Since the circle passes through point (2, 3)
$\therefore$ (2 - h)² + (3 - k)² = r² $\Rightarrow$ 4 + h² - 4h + 9 + k² - 6k = r²
$\Rightarrow$ h² + k² -4h - 6k + 13= $r^2........(3)$
From (2) and (3), we have
h² + k² -4h - 6k + 13 = h² + k²  + 2h - 2k + 2
$\Rightarrow$ -6h - 4k = -11 $\Rightarrow$ 6h + 4k = 11 .... (4)
Since the centre (h, k) of the circle lies on the line (x - 3y - 11 = 0)
$\therefore$ h - 3k - 11 = 0 $\Rightarrow$ h - 3k = 11 . . . (5)
Solving (4) and (5), we have,
$h = \frac{7}{2}$ and  $k = \frac{{ - 5}}{2}$
Putting these values of h and k in (3), we have
${\left( {\frac{7}{2}} \right)^2} + {\left( {\frac{{ - 5}}{2}} \right)^2} - \frac{{4 \times 7}}{2} - 6 \times \frac{{ - 5}}{2} + 13 = {r^2}$
$\Rightarrow \frac{{49}}{4} + \frac{{25}}{4} - 14 + 15 + 13 = {r^2} \Rightarrow {r^2} = \frac{{65}}{2}$
Thus equation of required circle is
${\left( {x - \frac{7}{2}} \right)^2} + {\left( {y + \frac{5}{2}} \right)^2} = \frac{{65}}{2}$$\Rightarrow {x^2} + \frac{{49}}{4} - 7x + {y^2} + \frac{{25}}{4} + 5y = \frac{{65}}{2}$
$\Rightarrow$ 4x² + 49 – 28x + 4y² + 25 + 20y = 130
$\Rightarrow$ 4x² + 4y² – 28x + 20y – 56 = 0
$\Rightarrow$ 4(x² + y² – 7x + 5y – 14) = 0
$\Rightarrow$ x² + y² – 7x + 5y – 14 = 0

The equation of the parabola takes the form x² = 4ay. Since it passes through
$\left(6, \frac{3}{100}\right)$ we have $(6)^{2}=4 a\left(\frac{3}{100}\right)$ , i.e., $a=\frac{36 \times 100}{12}=300 \mathrm{m}$ $\Rightarrow$ a = 300m
Let AB be the deflection of the beam which is  $\frac{1}{100} \mathrm{m}$ Coordinates of B are $\left(x, \frac{2}{100}\right)$
Therefore $x^{2}=4 \times 300 \times \frac{2}{100}=24$
i.e. $x=\sqrt{24}=2 \sqrt{6}$ m

length of the major axis = 2a = 6 units

the length of the minor axis = 2b = 4 units and

the eccentricity of the ellipse = $\frac{\sqrt{5}}{3}$.